Trigonometry/For Enthusiasts/Doing without Sine

The IdeaEdit

We know that:

\sin^2\theta = 1 - \cos^2\theta\,

So do we really need the \sin\, function?

Or put another way, could we have worked out all our interesting formulas for things like \cos( a+b )\, in terms just of \cos\, and then derived every formula that has a \sin\, in it from that?

The answer is yes.

We don't need to have one geometric argument for \cos(a+b)\, and then do another geometric argument for \sin(a+b)\,. We could get our formulas for \sin\, directly from formulas for \cos\,

Angle Addition and Subtraction formulasEdit

To find a formula for \cos(\theta_1+\theta_2)\, in terms of \cos\left( \theta_1\right) and \cos\left( \theta_2\right): construct two different right angle triangles each drawn with side c\, having the same length of one, but with \theta_1 \ne \theta_2, and therefore angle \psi_1 \ne \psi_2. Scale up triangle two so that side a_2\, is the same length as side c_1\,. Place the triangles so that side c_1\, is coincidental with side a_2\,, and the angles \theta_1\, and \theta_2\, are juxtaposed to form angle \theta_3=\theta_1+\theta_2\, at the origin. The circumference of the circle within which triangle two is embedded (circle 2) crosses side a_1\, at point g\,, allowing a third right angle to be drawn from angle \varphi_2 to point g\, . Now reset the scale of the entire figure so that side c_2\, is considered to be of length 1. Side a_2\, coincidental with side c_1\, will then be of length \cos\left( \theta_1\right), and so side a_1\, will be of length \cos\left( \theta_1\right) \cdot \cos\left( \theta_2\right) in which length lies point g\,. Draw a line parallel to line a_1\, through the right angle of triangle two to produce a fourth right angle triangle, this one embedded in triangle two. Triangle 4 is a scaled copy of triangle 1, because:

 (1) it is right angled, and 
 (2) \theta_4 + (\pi - \frac{\pi}{2} - \theta_1 - \theta_2) = \varphi_2 = \pi - \frac{\pi}{2} - \theta_2 \Rightarrow \theta_4 = \theta_1. 

The length of side  b_4\, is  cos(\varphi_2)cos(\varphi_4) = cos(\varphi_2)cos(\varphi_1) as  \theta_4\, = \theta_1\, . Thus point  g\, is located at length:

 cos(\theta_1+\theta_2) = cos(\theta_1)cos(\theta_2) - cos(\varphi_1)cos(\varphi_2), where \theta_1+\varphi_1 = \theta_2+\varphi_2 = \frac{\pi}{2}

giving us the "Cosine Angle Sum Formula".

Proof that angle sum formula and double angle formula are consistentEdit

We can apply this formula immediately to sum two equal angles:

    cos(2\theta) = cos(\theta+\theta) = cos(\theta)cos(\theta) - cos(\varphi)cos(\varphi)  = cos(\theta)^{2} - cos(\varphi)^{2}            (I)
    where  \theta+\varphi = \frac{\pi}{2} 

From the theorem of Pythagoras we know that:

    a^2 + b^2 = c^2\, 

in this case:

    cos(\theta)^2 + cos(\varphi)^2  = 1^2 
   \Rightarrow cos(\varphi)^2 = 1 - cos(\theta)^2 
   where  \theta+ \varphi = \frac{\pi}{2} 

Substituting into (I) gives:

   cos(2\theta)=cos(\theta)^2-cos(\varphi)^2
   = cos(\theta)^2 - (1 - cos(\theta)^2)\,
   = 2 cos(\theta)^2 - 1\,
   where \theta+\varphi = \frac{\pi}{4}

which is identical to the "Cosine Double Angle Sum Formula":

   2 cos(\delta)^2 - 1 = cos(2\delta)\,

Pythagorean identityEdit

Armed with this definition of the sin()\, function, we can restate the Theorem of Pythagoras for a right angled triangle with side c of length one, from:

    cos(\theta)^2 + cos(\varphi)^2 = 1^2  where \theta+\varphi = \frac{\pi}{2}

to:

    cos(\theta)^2 + sin(\theta)^2 = 1\,

We can also restate the "Cosine Angle Sum Formula" from:

 cos(\theta_1+\theta_2) = cos(\theta_1)cos(\theta_2) - cos(\varphi_1)cos(\varphi_2) where \theta_1+\varphi_1 = \theta_2+\varphi_2 = \frac{\pi}{2}

to:

 cos(\theta_1+\theta_2) = cos(\theta_1)cos(\theta_2) - sin(\theta_1)sin(\theta_2)\,

Sine FormulasEdit

The price we have to pay for the notational convenience of this new function sin()\, is that we now have to answer questions like: Is there a "Sine Angle Sum Formula". Such questions can always be answered by taking the cos()\, form and selectively replacing cos(\theta)^2\, by 1 - sin(\theta)^2\, and then using algebra to simplify the resulting equation. Applying this technique to the "Cosine Angle Sum Formula" produces:

   cos(\theta_1+\theta_2) = cos(\theta_1)cos(\theta_2) - sin(\theta_1)sin(\theta_2)\,
   \Rightarrow cos(\theta_1+\theta_2)^2 = (cos(\theta_1)cos(\theta_2) - sin(\theta_1)sin(\theta_2))^2\,
   \Rightarrow 1 - cos(\theta_1+\theta_2)^2 = 1 - (cos(\theta_1)cos(\theta_2) - sin(\theta_1)sin(\theta_2))^2\,
   \Rightarrow sin(\theta_1+\theta_2)^2     = 1 - (cos(\theta_1)^2cos(\theta_2)^2 + sin(\theta_1)^2sin(\theta_2)^2 - 2cos(\theta_1)cos(\theta_2)sin(\theta_1)sin(\theta_2))\,
   -- Pythagoras on left, multiply out right hand side
                     = 1 - (cos(\theta_1)^2(1-sin(\theta_2)^2) + sin(\theta_1)^2(1-cos(\theta_2)^2) - 2cos(\theta_1)cos(\theta_2)sin(\theta_1)sin(\theta_2))\,
                     -- Carefully selected Pythagoras again on the left hand side
                     = 1 - (cos(\theta_1)^2-cos(\theta_1)^2sin(\theta_2)^2 + sin(\theta_1)^2-sin(\theta_1)^2cos(\theta_2)^2 - 2cos(\theta_1)cos(\theta_2)sin(\theta_1)sin(\theta_2))\,
                     -- Multiplied out
                     = 1 - (1 -cos(\theta_1)^2sin(\theta_2)^2 -sin(\theta_1)^2cos(\theta_2)^2 - 2cos(\theta_1)cos(\theta_2)sin(\theta_1)sin(\theta_2))\,
                     -- Carefully selected Pythagoras 
                     = cos(\theta_1)^2sin(\theta_2)^2+sin(\theta_1)^2cos(\theta_2)^2 + 2cos(\theta_1)cos(\theta_2)sin(\theta_1)sin(\theta_2)\,
                     -- Algebraic simplification
                     =(cos(\theta_1)sin(\theta_2)+sin(\theta_1)cos(\theta_2))^2\,

taking the square root of both sides produces the "Sine Angle Sum Formula"

\Rightarrow sin(\theta_1+\theta_2) = cos(\theta_1)sin(\theta_2) + sin(\theta_1)cos(\theta_2)

We can use a similar technique to find the "Sine Half Angle Formula" from the "Cosine Half Angle Formula":

cos \left (\frac{\theta}{2} \right ) = \sqrt{\frac{1 + cos(\theta)}{2}}

We know that 1 - cos \left ( \frac{\theta}{2} \right ) ^2 = sin \left ( \frac{\theta}{2} \right ) ^2, so squaring both sides of the "Cosine Half Angle Formula" and subtracting from one:

 \Rightarrow 1 - cos \left ( \frac{\theta}{2} \right ) ^2 =       1 - \frac{1+cos(\theta)}{2}
 \Rightarrow     sin \left ( \frac{\theta}{2} \right ) ^2  =            \frac{1-cos(\theta)}{2}
  \Rightarrow     sin \left ( \frac{\theta}{2} \right ) = \sqrt{\frac{1-cos(\theta)}{2}}

So far so good, but we still have a \frac{cos(\theta)}{2} to get rid of. Use Pythagoras again to get the "Sine Half Angle Formula":

         sin\left (\frac{\theta}{2} \right ) = \sqrt{\frac{1 - \sqrt{1 - sin(\theta)^2}}{2}}

or perhaps a little more legibly as:

         sin \left ( \frac{\theta}{2} \right  ) = \sqrt{\frac{1}{2}}\sqrt{1 - \sqrt{1 - sin(\theta)^2}}
Last modified on 21 January 2011, at 19:36